Termination w.r.t. Q proof of TRS/nontermin/CSREx6_9

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, X1))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, X1))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

2ND₁(cons₂(X, X1)) -> 2ND₁(cons1₂(X, X1))
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, X1))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

2ND₁(cons₂(X, X1)) -> 2ND₁(cons1₂(X, X1))
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, X1))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, X1))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.